# Is there an autocorrelation function with a constant integral whose absolute value integral diverges?

Suppose a function $g:\mathbb{R}\rightarrow\mathbb{R}$ such that:

1. $|g(x)|\leq g(0)$;
2. $g(x)=g(-x)$, i.e. $g(x)$ is even;
3. $\int_{-\infty}^{\infty}g(x)dx=C$;
4. There exists a Fourier transform of $g(x)$, $\mathcal{F}(g(x))=G(\xi)$. Since $g(x)$ is even, $G(\xi)$ is real, even, and non-negative. Also, the aforementioned constant $C=G(0)$.

Here, my $g(x)$ is an autocorrelation function and $G(\xi)$ is the corresponding power-spectral density.

Given the conditions above, is it possible that:

$$\int_{-\infty}^{\infty}|g(x)|dx=\infty$$

I can't think of an example $g(x)$ where this happens, nor of a proof that such $g(x)$ does not exist. Any help?

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First of all, the fact that $g$ is even does not imply that $G(\xi)\ge0$; just think taht $-g$ is also even, and $\mathcal{F}(-g)=-G$.
The Fourier transform is usually defined for integrable functions $g$, that is, functions such that $\int_{-\infty}^{+\infty}|g(x)|\,dx<\infty$. I will assume that you mean that $$G(\xi)=\lim_{R\to\infty}\int_{-R}^{R}g(x)e^{ix\xi}\,dx\text{ exists.}$$ Let $$g(x)=\frac{\sin x}{x}\text{ if }x\ne0,\quad f(0)=1.$$ Then $g$ is even, $|g(x)|\le1=g(0)$ for all $x\in\mathbb{R}$, $\int_{-\infty}^{+\infty}|g(x)|\,dx=\infty$ and $$G(\xi)=c\,\chi_{[-1,1]}(x),$$ where $c$ is a constant (depending on the definition of the Fourier transform) and $\chi_A$ is tha characteristic function of $A\subset\mathbb{R}$. In this particular example, $G$ is in fact non-negative.
Thanks for catching my mistake on the non-negativity of $G(\xi)$ -- it's non-negative due to the other properties of power spectral density (one way to prove it is apply an arbitrary band-pass filter and show that PSD non-negative everywhere since power can't be negative). – M.B.M. Apr 7 '12 at 2:53
And thanks for you counterexample. For some reason the $\operatorname{sinc}(\cdot)$ function didn't come to my mind when I was thinking about this. – M.B.M. Apr 7 '12 at 2:54